I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 5.7 : Computing Definite Integrals
6. Evaluate the following integral, if possible. If it is not possible clearly explain why it is not possible to evaluate the integral.
\[\int_{1}^{2}{{\frac{1}{{7z}} + \frac{{\sqrt[3]{{{z^2}}}}}{4} - \frac{1}{{2{z^3}}}\,dz}}\]Show All Steps Hide All Steps
Start SolutionFirst we need to integrate the function.
\[\int_{1}^{2}{{\frac{1}{{7z}} + \frac{{\sqrt[3]{{{z^2}}}}}{4} - \frac{1}{{2{z^3}}}\,dz}} = \int_{1}^{2}{{\frac{1}{7}\frac{1}{z} + \frac{1}{4}{z^{\frac{2}{3}}} - \frac{1}{2}{z^{ - 3}}\,dz}} = \left. {\left( {\frac{1}{7}\ln \left| z \right| + \frac{3}{{20}}{z^{\frac{5}{3}}} + \frac{1}{4}{z^{ - 2}}} \right)} \right|_1^2\]Recall that we don’t need to add the “+c” in the definite integral case as it will just cancel in the next step.
Show Step 2The final step is then just to do the evaluation.
We’ll leave the basic arithmetic to you to verify and only show the results of the evaluation. Make sure that you evaluate the upper limit first and then subtract off the evaluation at the lower limit.
Here is the answer for this problem.
\[\int_{1}^{2}{{\frac{1}{{7z}} + \frac{{\sqrt[3]{{{z^2}}}}}{4} - \frac{1}{{2{z^3}}}\,dz}} = \left( {\frac{1}{7}\ln \left( 2 \right) + \frac{3}{{20}}\left( {{2^{\frac{5}{3}}}} \right) + \frac{1}{{16}}} \right) - \left( {\frac{1}{7}\ln \left( 1 \right) + \frac{2}{5}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{7}\ln \left( 2 \right) + \frac{3}{{20}}\left( {{2^{\frac{5}{3}}}} \right) - \frac{{27}}{{80}}}}\]Don’t forget that \(\ln \left( 1 \right) = 0\)! Also, don’t get excited about “messy” answers like this. They happen on occasion.